Magnetic Field of Four Parallel Wires

[Angie K.]

Homework Statement

Four long straight wires located at the corners of a square of side l = 27.8 cm carry equal currents I0 = 16 A perpendicular to the page as shown in the figure above. Determine the magnitude of B at the center of the square.

Homework Equations

B=mu0/2pi*I/r

The Attempt at a Solution

I know that using the equation above I will get the magnetic field. But does it matter that it's asking for the magnetic field at the center? Also when doing the x and y component, I am not sure how to find the angle. Is it just 90 because it's perpendicular?

 

[BvU]

Sort of. Again !

What angle are you referring to ? 90 degrees with respect to what ?

The magnetic field is linear in the sense that individual contributions can simply be added. But you have (as you do) to keep in mind that they are vectors, so you need some vector addition pizazz. Who knows you don't have to calculate anything or only one field strength if you do is smartly...

But it's best to befgin with the field at the center due to just 1 of the wires and then see how to continue

--

 

[Angie K.]

Angle with respect to length of wire and magnetic field. So the total magnetic field would be the sum of field of all 4 currents?

 

[BvU]

Yes, the angle wrt the wire is 90 degrees. So all four contributions are in the plane of the page.
But 90 degrees wrt "and magnetic field" is a little unclear to me. If not to say contradictory. What precisely do you mean ? Perhaps a drawing ?

And yes "the total magnetic field would be the sum of field of all 4 currents"

 

[Angie K.]

Yes, the angle wrt the wire is 90 degrees. So all four contributions are in the plane of the page.
But 90 degrees wrt "and magnetic field" is a little unclear to me. If not to say contradictory. What precisely do you mean ? Perhaps a drawing ?

And yes "the total magnetic field would be the sum of field of all 4 currents"

I think I just had the wrong approach with the angle wrt to the wire statement that I said.

The radius of the point in the middle of the square and each current in .1966 m

So I can use the equations for the x and y components to solve this problem?

x components would be B1x+B2x+B3x+B4x (magnetic field of the 4 currents summed up)
y components would be B1y+B2y+B3y+B4y (magnetic field of the 4 currents summed up)

And to find Bx of each current I would use -Bcos (theta)
and to find By of each current I would use +Bsin (theta)

Then the total B would just be the square root of Bx^2+By^2

Where B =mu0*I/(2pi*r)

Is that the right approach?

 

[BvU]

Well, it's a start. How do you take the directions (into/out of page) of the currents into account ?

And I still think you can make this easier for yourself by making a drawing; see how the four $\vec B$ add up.